I have an error in integrating to calculate the age of the Universe

[PeterDonis]

If this is wrong, can you tell me where my error is?

I already showed you a simpler way of proceeding, that results in a different integral than the one you are doing. See post #17. The integral I gave there gives the correct answer; if you are doing a different integral, that would explain why you are getting the wrong answer.

 

[Buzz Bloom]

Hi @PeterDonis:

I do not understand why the integrand I used is not equivalent to the integrand you posted.
Mine is
$$dt = \frac {da} {H_0 a \sqrt {M/a3+L}} = \frac {{\sqrt {a}} da} {H_0 \sqrt {M+L a^3}}$$
Yours is based on
$$dt = \frac {da} {H_0 \sqrt{M/a + La^2} }$$.
The difference is multiplying the numerator and denominator of your integrand with $\sqrt {a}$.

 

[George Jones]

$$dt = \frac {da} {H_0 a \sqrt {M/a3+L}} = \frac {{\sqrt {a}} da} {H_0 \sqrt {M+L a^3}}$$
I make the following substitutions.
${x = a^3,}$ ${a = x^{1/3}},$ ${da = dx / 3x^{2/3}},$

(I am not going to use $Q$.) The substitution that I prefer for this is
$$\frac{L}{M} a^3 = \sinh^2 u .$$
I make this substitution in order to make use of the identity $\cosh^2 u - \sinh^2 u = 1$. Differentiate:
$$\begin{align}
3\frac{L}{M} a^2 da &= 2\sinh u \frac{d}{du} \sinh u du \\
&= 2\sinh u \cosh u du \\
da &= \frac{2M}{3L} a^{-2} \sinh u \cosh u du
\end{align}$$
Substituting these into the integral gives
$$\begin{align}
\frac {{\sqrt {a}} da} {H_0 \sqrt {M + L a^3}} &= \frac {{\sqrt {a}} da} {H_0 \sqrt {M} \sqrt{1 + \frac{L}{M} a^3}} \\
&= \frac {{\sqrt {a}} \frac{2M}{3L} a^{-2} \sinh u \cosh u du} {H_0 \sqrt {M} \sqrt{1 + \sinh^2 u}} .
\end{align}$$
Now (if you want to proceed with this method), combine the $a$ s, and then replace them by something in terms of $u$ using my original substitution. Also, use the (hyperbolic) trig identity that I gave.

This should result in an expression for $t\left(a\right)$ that involves an inverse $\sinh$, and that, if desired, can reexpressed using $\ln$ s.

 

[PeterDonis]

Mine is
$$dt = \frac {da} {H_0 a \sqrt {M/a3+L}} = \frac {{\sqrt {a}} da} {H_0 \sqrt {M+L a^3}}$$

Not here it isn't:

$$ dt =({\frac {1} {3H_0 \sqrt L}}) ( {\frac {1} { \sqrt {x^2 + 2Qx}} } )$$

So which one are you integrating?

 

[Buzz Bloom]

Hi @PeterDonis:

I do not understand why the integrand I uses is not equivalent to the integrand you posted.
Mine is
$$dt = \frac {dx} {3H_0 \sqrt L \sqrt{x^2 + 2Qx} }$$.
Yours is based on
$$dt = \frac {dx} {3H_0 \sqrt L \sqrt{x^2 + 2Qx} }$$.
The process of integrating involves a sequence of substitutions which are intended to create an integrand which is finally directly integrated. The issue is whether the substitutions I made are OK, or whether they include an error. The final integrand is then integrated, and I have shown in Post #28 that differentiating the integration result leads to the integrand which was integrated. This demonstrates that the integration step was correct. The final part of the presentation in Post #19 is to use the desired variable values and arithmetically calculate the numerical value of t explained in Post #30. Note that the arithmetic conclusion is based on
$$\int_0^1 dt = {t(1) - t(0)}.$$

Regards,
Buzz

 

[PeterDonis]

I do not understand why the integrand I uses is not equivalent to the integrand you posted.

Just look at them. But first you have to get them right.

Mine is
$$dt = \frac {dx} {3H_0 \sqrt L \sqrt{x^2 + 2Qx} }$$.

Yes, that's what you posted earlier.

Yours is based on
$$dt = \frac {dx} {3H_0 \sqrt L \sqrt{x^2 + 2Qx} }$$.

No, it isn't. I think you mistakenly quoted yourself again. My integrand is this:

$$dt = \frac{1}{H_0} \frac{da}{\sqrt{\frac{M}{a} + L a^2}}$$

These are obviously not the same function. The integration variable being $x$ vs. $a$ is a cosmetic detail, the name of the variable doesn't matter, what matters is the functional form of the integrand. Since in both cases, the range of integration is $0$ to $1$, and the functions being integrated are different, obviously the numerical answers will be different. And, as you have already found, your integrand gives the wrong answer. Mine gives the right answer.

Evidently you made some mistake in the substitutions you made to get your integrand in $x$ from the original one. That's why you're getting the wrong answer.

 

[Buzz Bloom]

Hi @PeterDonis:

I did make an error in Post #35. Is is OK with you if I fix it with edits until I decide I have it right?

I think for any post I write I need to preface it with
"THIS IS A WORK IN PROGRESS,"
so if I make an error I can correct it before someone has responded to my error before I fixed it.

Regards,
Buzz

 

[PeterDonis]

I did make an error in Post #35. Is is OK with you if I fix it with edits until decide I have it right?

No. As has already been mentioned, once a post has responses, you should not do this. It is very confusing to other participants in the thread. Just make a follow-up post quoting what was in error from the previous one, and correcting it.

I think for any post I write I need to preface it with
"THIS IS A WORK IN PROGRESS,"
so if I make an error I can correct it before someone has responded to my error before I fixed it.

This basically means nobody can ever respond to your posts until we're sure you can't edit them any more. That's not going to be feasible for other thread participants.

There is a "preview" function in the post window; I suggest that you use it again and again to review what you are going to post before you post it. The site will save your draft post so there is no need to post it just to have it saved. You can keep a post in draft for a while while you edit, preview, edit, preview, etc. until you are confident it is as good as you can get it, then post it.

 

[PeterDonis]

I did make an error in Post #35. Is is OK with you if I fix it with edits until decide I have it right?

Rather than go round and round with this, why don't you just respond to what I said in post #36?

 

[Buzz Bloom]

Hi @PeterDonis:

Well, I finally completed #19, and now have shown the process that produces a correct numerical result for the integral.

Regarding #36, your equation

$$dt = \frac {1}{H_0} {\frac {da} { \sqrt {M/a+L a^2}}} $$

becomes

$$dt = \frac {{\sqrt {a}} da} {H_0 \sqrt {M+L a^3}}$$

by multiplying your numerator and denominator by $1/{\sqrt {a}}$.

Then I substitute $a^{1/3} = x$ and take L outside of the sqrt. Both x and a have the integrations limits of 0 and 1. I also define the constant Q = M/2L.

My error (which I fixed) was not in the form of the equation, but in my calculations of the numerical arithmetic.

Thank you for helping me get used to LaTeX.

Regards,
Buzz

 

[Buzz Bloom]

Hi @George Jones:

Re #33, my intent was to use the equation approach I presented (which you quoted in #33) to obtain a numerical result of about 13.8 giga-years. I finally was successful in Post #19. I was not trying to find a different way to integrate. Thank you for your interest.

Regards,
Buzz